Suppose X is the maximum number of consecutive heads in five tosses of a coin, which can be anything between 1 and 5. Then P(X=3) is equal to______.

To solve for \( P(X = 3) \), we need to determine the probability that the maximum number of consecutive heads in five tosses of a fair coin is exactly 3.

### Step 1: Total Possible Outcomes
The total number of outcomes when tossing a fair coin 5 times is:
\[
2^5 = 32
\]
Each sequence consists of heads (H) and tails (T).

### Step 2: Count of Sequences where Maximum Consecutive Heads is 3
To satisfy \( X = 3 \), we need sequences where:
- There is at least one sequence of exactly 3 consecutive heads.
- No sequence of heads contains more than 3 consecutive heads.

#### Cases to Consider
We will count the valid sequences that have exactly 3 consecutive heads.

1. **Three heads at the start:**
   The sequence is \( HHH \) followed by two additional flips. To ensure the maximum is exactly 3, at least one of the remaining flips must be tails.
   - Possible sequences: \( HHH TT, HHH TH, HHH HT \)
   - Count: 3

2. **Three heads in the middle:**
   The sequence is \( THHH \) followed by one additional flip or one flip before \( HHH \).
   - Possible sequences: \( THHH T, THHH H \) or \( T HHH T, H HHH T \) with exactly 3 heads.
   - Count: 4 (2 from each side)

3. **Three heads at the end:**
   The sequence starts with two flips followed by \( HHH \). To ensure the maximum is exactly 3, at least one of the first two flips must be tails.
   - Possible sequences: \( TT HHH, TH HHH, HT HHH \)
   - Count: 3

Thus, the total number of valid sequences where the maximum number of consecutive heads is exactly 3 is:
\[
3 + 4 + 3 = 10
\]

### Step 3: Probability
Now, the probability is the ratio of favorable outcomes (10) to the total number of outcomes (32):
\[
P(X = 3) = \frac{10}{32} = \frac{5}{16}
\]

Thus, the probability \( P(X = 3) \) is \( \frac{5}{16} \).

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Would you like further details or clarification on any step?

Here are 5 related questions for further exploration:
1. What is \( P(X = 4) \)?
2. How does the probability change if the number of coin tosses increases to 6?
3. How can you generalize this method to find \( P(X = k) \) for any number of tosses?
4. What is the expected value of \( X \) for 5 coin tosses?
5. What is the probability that there are no consecutive heads in 5 tosses?

**Tip**: In problems involving consecutive events (like heads in coin tosses), using pattern recognition and breaking the problem into smaller cases simplifies the solution.

Find the probability that the maximum number of consecutive heads in five tosses of a coin is exactly 3. This detailed step-by-step solution involves concepts of probability, combinatorics, and counting sequences. Suitable for students in Grades 10-12.

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